Recipe

Run the bisection method

Use when: you have a bracket $[a,b]$ (f(a), f(b) of opposite sign) and need a root to a stated tolerance

Goal: an approximation within the required tolerance of a root, with a guaranteed error bound at every step

Taught in The Bisection Method

Quick recipe

Confirm the bracket is valid
Compute the midpoint and evaluate $f$ there
Replace whichever endpoint shares $f(c)$ 's sign
Repeat until the stopping tolerance is met

1
Evaluate $f(a)$ and $f(b)$ and check that they have opposite signs. If they don't, bisection has nothing to work with — go back and find a real bracket first.

Example

For $f(x)=x^3-x-2$ on $[1,2]$ : $f(1)=1-1-2=-2$ and $f(2)=8-2-2=4$ . Opposite signs — this is a valid bracket.

Why: every later step depends on the bracket actually containing a sign change; bisection has no way to detect or recover from starting on an invalid one.
2
Compute $c=\dfrac{a+b}{2}$ and evaluate $f(c)$ .

Example

$c=\dfrac{1+2}{2}=1.5$ . $f(1.5)=1.5^3-1.5-2=3.375-1.5-2=-0.125$ .

Why: $c$ is the single new piece of information each step needs — it's the point whose sign decides which half of the bracket survives. (Section 8 revisits this exact step with a smarter candidate point than the plain midpoint, keeping every other part of this recipe unchanged.)
3
Compare the sign of $f(c)$ to the sign of $f(a)$ . If they match, set $a\leftarrow c$ (the root is in the other half, $(c,b)$ ). If $f(c)$ matches $f(b)$ 's sign instead, set $b\leftarrow c$ .

Example

$f(1)=-2$ is negative, and $f(1.5)=-0.125$ is also negative — they match, so $a\leftarrow1.5$ . The new bracket is $[1.5,2]$ , half the width of $[1,2]$ .
- $f(c)=0$ exactly — stop immediately; $c$ is a root, not just an approximation of one.
- $f(c)$ matches neither sign because $f(c)=0$ landed exactly between — the exact-root case above already covers this; there is no third outcome.
Why: keeping the half whose endpoints still disagree in sign is what guarantees, by the same Intermediate Value Theorem argument that found the original bracket, that a root still lies inside the new one.
4
Go back to Step 2 with the new bracket, and keep going until $|b-a|$ (or $|f(c)|$ , depending on which stopping rule was asked for) is below the required tolerance.

Example

Continuing from $[1.5,2]$ : $c=1.75$ , $f(1.75)\approx1.609>0$ , matches $f(2)=4>0$ , so $b\leftarrow1.75$ , giving $[1.5,1.75]$ . Next, $c=1.625$ , $f(1.625)\approx0.666>0$ , again matches $f(b)$ , giving $[1.5,1.625]$ . After 8 steps total the bracket has narrowed to $[1.515625,\,1.5234375]$ , width $\approx0.0078$ , closing in on the true root $x\approx1.52138$ .
- stopping on bracket width $|b-a|<\varepsilon$ — guarantees the root is within $\varepsilon$ of either endpoint, by the theorem above.
- stopping on $|f(c)|<\varepsilon$ — guarantees $c$ makes $f$ small, which is a reasonable proxy for closeness to a root but not identical to it (a very steep $f$ can have small $|f(c)|$ while $c$ itself is still relatively far from the root).
⚡ Shortcut

The convergence bound also runs in reverse: to hit a target tolerance $\varepsilon$ , solve $\dfrac{b-a}{2^n}<\varepsilon$ for $n$ directly, before running a single step, instead of stepping and checking after each one: $n > \log_2\!\left(\frac{b-a}{\varepsilon}\right)$ For the bracket $[1,2]$ (width $1$ ) and $\varepsilon=0.0001$ : $n>\log_2(1/0.0001)=\log_2(10{,}000)\approx13.29$ , so $n=14$ steps are needed — checked directly: $1/2^{13}\approx0.000122$ is still too wide, but $1/2^{14}\approx0.000061$ clears the tolerance. Knowing the exact iteration count ahead of time is useful for budgeting how expensive a bisection run will be, especially when each evaluation of $f$ is costly (a simulation, an experiment) — no other method in this chapter allows this kind of advance guarantee, because none of them shrinks by a fixed, function-independent factor every step.

Why: each pass through this step is exactly what the convergence theorem above is a guarantee about — the bracket width after $n$ passes is provably $(b-a)/2^n$ , so continuing is what actually delivers on that promise.

Common mistakes

Starting bisection on an interval that was never confirmed to be a valid bracket — the method runs without complaint but converges to nothing meaningful.
Replacing the wrong endpoint — a slip here (setting $b\leftarrow c$ when the sign match was actually with $f(b)$ , not $f(a)$ ) throws away the half that actually contains the root.
Re-evaluating $f$ at both $a$ and $b$ every step. Only $f(c)$ is new information each step; whichever endpoint survives keeps the function value it already had, computed on a previous step.
Stopping after a fixed number of iterations chosen by guesswork instead of checking the actual tolerance required — the shortcut above computes the right number directly instead of guessing.
Treating a small $|f(c)|$ and a small bracket width as automatically the same guarantee — they usually track together but aren't identical (see the case note above).